Particles and Fields – a neverending story

It is often said that the field operator [tex]\hat{\phi}(x)[/tex] of some free field theory, e.g. Klein-Gordon or Dirac, acting on the vacuum state, creates "one particle localized at x" or "in a position eigenstate at x".

For example, from the Fourier expansion of the real Klein-Gordon field operator, we get

where [tex]\mathrm{d}^3 \tilde{p} \equiv \mathrm{d}^3 p/2p^0[/tex] is the Lorentz covariant measure and [tex]f_p(x)[/tex] are the plane wave solutions to the Klein-Gordon equation.

Now, this state could be regarded as a localized one-particle state if [tex]f_p^*(x)[/tex] is the ordinary QM position space wave-function of the particle in momentum eigenstate [tex]|\mathbf{p}\rangle[/tex], namely [tex]f_p^*(x)=\langle \mathbf{p} |\mathbf{x} \rangle[/tex]. Writing [tex]a^\dagger(p) |0\rangle \equiv |\mathbf{p}\rangle[/tex], the above equation would turn into

which would somehow be related to [tex]|\mathbf{x}\rangle[/tex] since the first part resembles an identity operator. But two problems occur:

1. What is [tex]|\mathbf{x}\rangle[/tex]? While I can define the state of "one particle with momentum p" to be [tex]a^\dagger(p) |0\rangle \equiv |\mathbf{p}\rangle[/tex], I have no idea how a state like [tex]|\mathbf{x}\rangle[/tex] can canonically arise in a QFT. My only idea is to promote the equation

[tex]f_p^*(x)=\langle \mathbf{p} |\mathbf{x} \rangle[/tex]

to a defining axiom for [tex]|\mathbf{x}\rangle[/tex], that is, postulating that the free field solutions to the real Klein Gordon equation can be interpreted as one-particle wave functions in the ordinary QM sense. But this would mean you have to add a new axiom relating QM and QFT?!

2. I have some doubts that I can regard [tex]\int \mathrm{d}^3 \tilde{p} \: |\mathbf{p} \rangle \langle \mathbf{p} |[/tex] as an identity operator because of the covariant measure.

The problem with the states
[tex]|\mathbf{x}\rangle = \hat{\phi}(\mathbf{x},0)|0\rangle[/tex]
is that they are not orthogonal.
Namely,
[tex]\langle\mathbf{x}'|\mathbf{x}\rangle \neq \delta^3(\mathbf{x}-\mathbf{x}')[/tex]
Note, however, that they become orthogonal in the non-relativistic limit.

is used to obtain the transformation behavior of the field operator from the transformation behavior of the single-particle wave function. In a way, I use the representation theory of the Poincaré group to construct QM single-particle basis states and relate their transformation behavior to field operators of QFT by the above equation. Is this right?

But I don't get the connection; I think this is a subtle point: from the viewpoint of second quantization, everything is clear because one explicitly constructs the possibility to create and annihilate particles, so the above equation just follows mathematically.

But from the viewpoint of field quantization, I never introduced single-particle wave functions, I just applied the QM axioms to fields so that there is no obvious interpretation of the function which is represented by the above matrix element.
Does this make any sense?

There is no generally accepted answer to your question.
Nevertheless, a discussion that might be helpfull is provided in Secs. 8 and 9 ofhttp://xxx.lanl.gov/abs/quant-ph/0609163 [Found. Phys. 37 (2007) 1563]

For example, it says that a particle cannot be localized in a region smaller than its Compton length. However, if it was true, it would imply that a massless photon could not be localized in any region smaller than its Compton length. On the other hand, the position of the photon can be measured by certain accuracy.

Next, it says that in relativistic QM there is no such thing as a 2, 1, or 0 particle state. But for free relativistic fields, such things certainly do exist. Moreover, a 1 and 0 particle states are stable even for interacting particles.

1. Position of a photon? But you know that a photon has no rest frame...

2. I think every QFT text starts with this observation. What is wrong with it?

1. So what? Take for example a classical photon. It has no rest frame too. Yet, at any time it has a definite position. The position, of course, changes with time, but it does not mean that the position does not exist.

2. For example, what is the state |0>, if not the state with exactly zero number of particles? And no, not every QFT text starts with this observation.

For example, it says that a particle cannot be localized in a region smaller than its Compton length.

This is a point related to my question. What does particle localization mean in QFT? I am able to superpose some one-particle momentum eigenstates. But to determine the position of the "quantum object" thus obtained I need a position operator which does not exist. I cannot see more than a formal analogy to second quantization where it is just a mathematical implication that the fourier transform of the superposition weight function is the one-particle wave function where an ordinary QM position operator can act on.

It depends first on how you define "particle". I.e., which group you start with for which
you find the unitary irreducible representations. The Poincare algebra doesn't come
with a position generator, so some people advocate constructing one (Newton-Wigner)
from the enveloping ring. Other people are unimpressed with such "schtick". :-)

The x parameters in orthodox QFT (and Minkowski space in general) just serve as
a representation space for the Poincare group. The vacuum state has deterministic
energy-momentum of 0, and therefore cannot be a localised state in the usual sense.
Some people have suggested that this fact is at the core of the Reeh-Schlieder
paradox (which "proves" that fields over "there" can be recovered entirely from
fields "here" -- where "here" and "there" mean spacelike-separated regions).

Here's some fairly recent papers that debate these still-controversial points.

Abstract:
The Reeh-Schlieder theorem asserts the vacuum and certain other states to be spacelike
superentangled relative to local fields. This motivates an inquiry into the physical status
of various concepts of localization. It is argued that a covariant generalization of Newton-
Wigner localization is a physically illuminating concept. When analyzed in terms of
nonlocally covariant quantum fields, creating and annihilating quanta in Newton-Wigner
localized states, the vacuum is seen to not possess the spacelike superentanglement that
the Reeh-Schlieder theorem displays relative to local fields, and to be locally empty as
well as globally empty. Newton-Wigner localization is then shown to be physically
interpretable in terms of a covariant generalization of the center of energy, the two
localizations being identical if the system has no internal angular momentum. Finally,
some of the counterintuitive features of Newton-Wigner localization are shown to have
close analogues in classical special relativity.

Abstract:
Many of the “counterintuitive” features of relativistic quantum
field theory have their formal root in the Reeh-Schlieder theorem,
which in particular entails that local operations applied to the vacuum
state can produce any state of the entire field. It is of great interest
then that I.E. Segal and, more recently, G. Fleming (in a paper entitled
“Reeh-Schlieder meets Newton-Wigner”) have proposed an alternative
“Newton-Wigner” localization scheme that avoids the Reeh-Schlieder
theorem. In this paper, I reconstruct the Newton-Wigner localization
scheme and clarify the limited extent to which it avoids the counterin-
tuitive consequences of the Reeh-Schlieder theorem. I also argue that
there is no coherent interpretation of the Newton-Wigner localization
scheme that renders it free from act-outcome correlations at spacelike
separation.

3) DeBievre, "Where's that quantum"? Available as math-ph/0607044

Abstract:
The nature and properties of the vacuum as well as the meaning
and localization properties of one or many particle states have at-
tracted a fair amount of attention and stirred up sometimes heated
debate in relativistic quantum field theory over the years. I will review
some of the literature on the subject and will then show that these is-
sues arise just as well in non-relativistic theories of extended systems,
such as free bose fields. I will argue they should as such not have given
rise either to surprise or to controversy. They are in fact the result
of the misinterpretation of the vacuum as “empty space” and of a too
stringent interpretation of field quanta as point particles. I will in par-
ticular present a generalization of an apparently little known theorem
of Knight on the non-localizability of field quanta, Licht’s character-
ization of localized excitations of the vacuum, and explain how the
physical consequences of the Reeh-Schlieder theorem on the cyclic-
ity and separability of the vacuum for local observables are already
perfectly familiar from non-relativistic systems of coupled oscillators.

Thanks very much for the detailed reply! I will have a look at some of these papers. But I'm still confused why in standard textbooks on QFT, say, a Gaussian superposition of momentum eigenstates is called localized (blurred around a certain point x) in space. Which reasoning is behind this?

say, a Gaussian superposition of momentum eigenstates is called localized
(blurred around a certain point x) in space. Which reasoning is behind this?

IMHO, I think that's just what people would like to think. I.e., we'd like to
think that the x's we use when constructing orthodox QFT do indeed correspond
meaningfully to points in physical space. But one must delve into the subtleties
of position operators, understand the Reeh-Schlieder theorem, etc, (and maybe
also Haag's theorem), before it becomes apparent that something is deeply
puzzling about this whole subject.

Of course, such delving is clearly not necessary to calculate the stunningly
accurate predictions of QFT, hence most people don't worry very much about it.

Srednicki for example, Chapter 5 (LSZ formula), Eqs. (5.6) and (5.7). He speaks of creating a particle localized in momentum and position space. (A draft of the book is available at http://www.physics.ucsb.edu/~mark/qft.html" [Broken].)

Okay, thanks to you all for your answers! I think I know now what to read to get on...

Strangerep, thanks for posting those references. I'll definitely take a closer look at the last one and maybe the other two as well some time soon.

You seem to know this stuff pretty well already, so I'd like to ask you specifically about something that I have believed to be true until now. (Right now I'm really confused. I have e.g. never heard of the Reeh-Schlieder theorem before). I've been saying things like this:

The state of a photon is in general a superposition of states with different momenta. Let's ignore other degrees of freedom and express this as

[tex]\int d^3p f(\vec p)a^\dagger(\vec p)|0\rangle[/tex]

where [itex]a^\dagger(\vec p)[/itex] is the creation operator that creates a one-particle state with momentum p when it acts on the vaccum. The Fourier transform of f can be interpreted as an ordinary wave function.

Is this completely false? I guess I always thought that since the f above is pretty much the same in relativistic QM as in non-relativistic QM, its Fourier transform should also be pretty much the same as in non-relativistic QM.

Is this completely false? I guess I always thought that since the f above is pretty much the same in relativistic QM as in non-relativistic QM, its Fourier transform should also be pretty much the same as in non-relativistic QM.

The right questions are: Can this Fourier transform be interpreted as the probability amplitude? In other words, is such a probability conserved? And is it defined in a relativistic covariant way?
See e.g. the Appendix inhttp://xxx.lanl.gov/abs/0804.4564 [not yet published]

The right questions are: Can this Fourier transform be interpreted as the probability amplitude?

If I get this right, this is the same as asking whether you can interpret a first-quantized field as a particle system obtained by second quantization which obviously is done without ever asking for reasons.

Strangerep, thanks for posting those references. I'll definitely take a closer look at the last one and maybe the other two as well some time soon.
You seem to know this stuff pretty well already, so [...]

(Urgle!?) No, I'm not an expert. I'm aware that the various puzzles exist, but I
don't know how to resolve them satisfactorily. (Actually, I don't think
anyone does - hence the ongoing debate.)